What is the area of the region between the graphs of $f(x)=\dfrac{4}{x}$, $g(x)=4$, and $x=4$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $12-4\ln(3)$ (Choice B) B $12-4\ln(4)$ (Choice C) C $24-4\ln(3)$ (Choice D) D $24-4\ln(4)$
Visualizing the area We sketch the graphs of $f$ and $g$ first. ${1}$ ${2}$ ${3}$ ${4}$ ${2}$ ${4}$ ${6}$ ${8}$ ${10}$ ${12}$ ${14}$ $f$ $g$ $y$ $x$ From the graph, it appears that $g(x)\ge f(x)$ between the point where the graphs intersect and $x=4$. From this we are looking to evaluate: $ \int_{a}^{4}\left( g(x)-f(x) \right)\,dx$ where $a$ is the $x$ -coordinate of the point of intersection. Finding the $x$ -coordinate of the intersection point We can find the $x$ -coordinate of the point of intersection by setting the functions equal to each other and solving the resulting equation. $\begin{aligned} f(x)&=g(x) \\\\ \dfrac{4}{x}&=4\\\\ 4 &=4x\\\\ 1 &=x \end{aligned}$ The graphs intersect where $x=1$. Setting up the definite integral Thus, the area of the shaded region pictured above is given by: $ \int_{1}^{4}\left(4-\left(\dfrac{4}{x}\right)\right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{1}^{4} \left( 4- \left(\dfrac{4}{x} \right) \right) \,dx \\\\ &= 4x-4\ln|x|~\Bigg|_{1}^{4} \\\\ &= \left( 16-4\ln|4| \right) - \left( 4-0 \right) \\\\ &= 12-4\ln(4) \end{aligned}$ Answer The area is $12-4\ln(4)$ square units.